Problem: What is the extraneous solution to these equations? $\dfrac{x^2 + 5x}{x + 8} = \dfrac{19x - 40}{x + 8}$
Answer: Multiply both sides by $x + 8$ $ \dfrac{x^2 + 5x}{x + 8} (x + 8) = \dfrac{19x - 40}{x + 8} (x + 8)$ $ x^2 + 5x = 19x - 40$ Subtract $19x - 40$ from both sides: $ x^2 + 5x - (19x - 40) = 19x - 40 - (19x - 40)$ $ x^2 + 5x - 19x + 40 = 0$ $ x^2 - 14x + 40 = 0$ Factor the expression: $ (x - 10)(x - 4) = 0$ Therefore $x = 10$ or $x = 4$ The original expression is defined at $x = 10$ and $x = 4$, so there are no extraneous solutions.